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3 years ago

PLEASE PLEASE HELP!! A pipe drips at 155mL/day- convert this rate to gallons/day (1gal=3.785L)

Chemistry

1 answer:

ASHA 777 [7]3 years ago

7 0

1 mL = $10^{-3}$ L

so $155$ mL =$155\times10^{-3}$ L

Now, $1$ gallon = $3.785$ L , or $1$ L= $\frac1{3.785}$ gallons

so $155\times10^{-3}$ L= $155\times10^{-3}\times \frac{1}{3.785}=40.95\times 10^{-3}$ gallons

and this is for each day, so just divide by 1 (day) to get the ratw gallons/day

i.e. $40.95\times 10^{-3}$ gal/day

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1- 3 Moles of CO2

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Products form based on limiting reagent (384g O2) :

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The limiting reagent is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, because the reaction cannot continue without it.

If I have 132,33 g of C3H8 and 384 g O2 we can calculate:

For 44 g of CH3H8 ⇒160 g of O2

With 132,33 g of CH3H8 ⇒ ×= 481,2 g of O2

(132,33×160÷44=481,2)

As this amount exceeds the quantity of O2 that we have, we can assume that the 384 g O2 will be totally consumed.

Calculations of the products formed in base of quantity of O2 (limiting reagent):

160 g O2 ⇒ 132 g CO2

384 g O2⇒ × = 316,8 g CO2

(384×132÷160= 316,8)

160 g O2 ⇒ 72 g H2O

384 g O2⇒ × = 172,8 g H2O

(384×72÷160= 172,8)

Calculations of the products formed in base of quantity of C3H8 (excess reagent):

44 g C3H8 ⇒ 132 g CO2

132,33 g C3H8 ⇒ × = 396,99 g CO2

(132,33×132÷44=396,99)

44 g C3H8 ⇒ 72 g H2O

132,33 g C3H8⇒ × = 216,54 g H2O

(132,33×72÷44= 216,54)

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